JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A loop \(ABCDEFA\) of straight edges has six corner points \(\mathrm{A}(0,0,0), \mathrm{B}(5,0,0),\)\(\mathrm{C}(5,5,0) ,\mathrm{D}(0,5,0), \mathrm{E}(0,5,5)\) and \(\mathrm{F}(0,0,5) .\) The magnetic field in this region is \(\overrightarrow{\mathrm{B}}=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{k}}) \;\mathrm{T}\). The quantity of flux through the loop \(\mathrm{ABCDEFA}(\text { in Wb })\) is
- A \(169\)
- B \(200\)
- C \(196\)
- D \(175\)
Answer & Solution
Correct Answer
(D) \(175\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{A}}_{\mathrm{ABCD}}=25 \hat{\mathrm{k}}\) \(\overrightarrow{\mathrm{A}}_{\mathrm{ADEF}}=25 \hat{\mathrm{i}}\) \(\overrightarrow{\mathrm{A}}_{\mathrm{net}}=25 \hat{\mathrm{i}}+25 \hat{\mathrm{k}}\)…
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